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Consider the following set of processes that need to be scheduled on a single CPU. All the times are given in milliseconds.

$$\begin{array}{|c|c|c|} \hline \textbf{Process Name} & \textbf{Arrival Time} & \textbf{Execution Time} \\\hline \text{A} & 0 & 6 \\ \text{B} & 3 & 2 \\ \text{C} & 5 & 4 \\ \text{D} & 7 & 6 \\ \text{E} & 10 & 3 \\\hline \end{array}$$

Using the *shortest remaining time first* scheduling algorithm, the average process turnaround time (in msec) is ____________________.

+21 votes

Best answer

+4 votes

**Solution:** Let the processes be A, ,C,D and E. These processes will be executed in following order. Gantt chart is as follows:First 3 sec, A will run, then remaining time A=3, B=2,C=4,D=6,E=3 Now B will get chance to run for 2 sec, then remaining time. A=3, B=0,C=4,D=6,E=3 Now A will get chance to run for 3 sec, then remaining time. A=0, B=0,C=4,D=6,E=3 By doing this way, you will get above gantt chart. Scheduling table:As we know, turn around time is total time between submission of the process and its completion. i.e turn around time=completion time-arrival time. i.e. TAT=CT-AT Turn around time of A = 8 (8-0) Turn around time of B = 2 (5-3) Turn around time of C = 7 (12-5) Turn around time of D = 14 (21-7) Turn around time of E = 5 (15-10) Average turn around time is (8+2+7+14+5)/5 = 7.2. Answer is 7.2.

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